Question: $\text B = \left[\begin{array}{rr}5 & -2 \\ -1 & 5\end{array}\right]$ and $\text A = \left[\begin{array}{rr}2 & 0 \\ 0 & 2\end{array}\right]$. Let $\text {H = BA}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{B}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{5} & {-2} \\ -1 & 5\end{array}\right]\left[\begin{array}{rr} {2} & 0 \\ {0} & 2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(5,-2)\cdot(2,0)\\\\ &=5 \cdot 2 - 2\cdot 0\\\\ &=10 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot 0 - 2\cdot 2 = -4$ (Choice B) B $-1 \cdot 0 + 5 \cdot 2 = 10$ (Choice C) C $-1 \cdot 2 + 5\cdot 0 = -2$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}= \left[\begin{array}{rr}10 & -4 \\ -2 & 10\end{array}\right]$